Which of the following aqueous solutions has the highest [OH– ]? A. A solution with a pH of 3.0 B. 1 × 10–4 M solution of HNO3 C. A solution with a pOH of 12.0 D. Pure water E. 1 × 10–3 M solution of NH4Cl

The solution that has the highest [OH-] is pure water.

For the solution of pH 3.0;

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 3.0 = 11.0

Since;

pOH = -log[OH-]

[OH-] = Antilog (-pOH)

[OH-] = Antilog (-11.0)

[OH-] = 1 × 10^-11 M

For the 1 × 10–4 M solution of HNO3;

[OH-] [H^+] = 1 × 10–14

[OH-] = 1 × 10–14/[H^+]

[OH-] =  1 × 10–14/ 1 × 10–4

[OH-] = 1 × 10–10 M

For the solution of pOH = 12

[OH-] = Antilog (-pOH)

[OH-] =  Antilog (-12)

[OH-] =  1 × 10–12 M

Pure water has OH- concentration of   1 × 10–7 M

For 1 × 10–3 M solution of NH4Cl;

Since NH4Cl yields the acid HCl in solution;

[H+] = 1 × 10–3 M

[OH-] =  1 × 10–14/ 1 × 10–3 M

[OH-] = 1 × 10–11

The solution that has the highest [OH-]  is  pure water..

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D. Pure water; [OH-] = 1e-7

Explanation:

The greater the concentration, the greater the basicity.

A. A solution with a pH of 3.0

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 3 = 11

pOH = -log[OH-]

[OH-] = Antilog(-pOH)

[OH-] = Antilog(-11)

[OH-] = 1e-11

B. 1 × 10–4 M solution of HNO3

pH = -log[H+]

log10{0.0001} = log10(10−4)=−4, by definition of the log function.

Thus pH = 4,

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 4 = 10

pOH = -log[OH-]

[OH-] = Antilog(-pOH)

[OH-] = Antilog(-10)

[OH-] = 1e-10

C. A solution with a pOH of 12.0

pOH = -log[OH-]

[OH-] = Antilog(-pOH)

[OH-] = Antilog(-12)

[OH-] = 1e-12

D. Pure water

Pure water is considered to neutral and the hydronium ion concentration is 1.0 x 10-7 mol/L which is equal to the hydroxide ion concentration.

[OH-] = 1e-7

E. 1 × 10–3 M solution of NH4Cl

NH4Cl dissolves in solution to form ammonium ions NH+4 which act as a weak acid by protonating water to form ammonia, NH3(aq) and hydronium ions H3O+(aq):

NH+4(aq) + H2O(l) → NH3(aq) + H3O+(aq)

Ka × Kb = 1.0×10−14 assuming standard conditions.

So, Ka(NH+4) = 1.0×10−14 / 1.8×10−5 = 5.56×10−10

Plug in the concentration and the Ka value into the expression:

Ka = [H3O+] × [NH3] / [NH+4]

5.56×10−10 ≈ [H3O+] × [NH3] / [0.001]

5.56×10−13 = [H3O+]^2

(as we can assume that one molecule hydronium must form for every one of ammonia that forms. Also, Ka is small, so x≪0.1.)

[H3O+] = 7.46×10−7

pH=−log[H3O+]

pH = −log(7.45×10−6)

pH ≈ 6.13

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 6.13 = 7.87

pOH = -log[OH-]

[OH-] = Antilog(-pOH)

[OH-] = Antilog(-7.87)

[OH-] = 1.345e-08

i think a, hope it .

explanation:

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.